Question

If $\left|\vec{a}\right|=4,\left|\vec{b}\right|=2$ and the angle between $\vec{a}$ and $\vec{b}$ is $\pi /6$ then ${\left(\vec{a}\times \vec{b}\right)}^2$ is equal to _______.

Answer 1

Answer: 16

We have
$\vec{a}\cdot \vec{b}=\left|\vec{a}\right|\left|\vec{b}\right|cos\frac{\pi }{6}$
$=4\times 2\times \frac{\sqrt{3}}{2}=4\sqrt{3}$
Now, ${\left(\vec{a}\times \vec{b}\right)}^2+{\left(\vec{a}\cdot \vec{b}\right)}^2=a^2{b}^2$
$\Rightarrow {\left(\vec{a}\times \vec{b}\right)}^2=48=16\times 4$
$\Rightarrow {\left(\vec{a}\times \vec{b}\right)}^2=16$